Question 310761
let x , x+2 , and x+4 be the integers


(x+4)^2 + 65 = x^2 + (x+2)^2


x^2 + 8x + 81 = 2x^2 + 4x + 4


0 = x^2 - 4x - 77


0 = (x-11)(x+7)


x = -7 (the given solution)


x = 11


so the numbers are  11 , 13 , 15